Question

How do you graph `y=(4-x)/(5x^2-4x-1)` using asymptotes, intercepts, end behavior?

Answer 1

The vertical asymptotes are `x=-1/5` and `x=1`
The horizontal asymptote is `y=0`
There is no slant asymptote:
The intercepts are `(4,0) and (0,-4)`

Explanation:

Let's factorise the denominator
`5x^2-4x-1=(5x+1)(x-1)`
As you cannot divide by `0`, the vertical asymptotes are
`x=-1/5` and `x=1`
As the degree of the numerator is `<` the degree of the denominator, there is no slant asymptote.

`lim_(x->-oo)y=lim_(x->-oo)-1/(5x)=0^+`

`lim_(x->+oo)y=lim_(x->+oo)-1/(5x)=0^-`

So `y=0` is a horizontal asymptote
The intercepts are
x-axis when `y=0` `=>` `x=4`

and y-axis when `x=0` `=>` `y=-4`
graph{(4-x)/(5x^2-4x-1) [-20.27, 20.27, -10.15, 10.15]}